On two recurrence formulas for two kinds of identities of Riemann Zeta function
TL;DRAbstract
For any complex s, let ζ(s) denote Riemann Zeta function. We have ζ(s)=sum from n=1 to ∞ (1/n~s) when Re(s)>1. Now we define A(n,k,l)=sum from α_1+α_2+……+α_k=n to ((α_1α_2…α_k)~1ζ(2α_1)ζ(2α_2)…ζ(2α_k)), where n≥k is a positive integer, α_+α_2+…α_k=n denotes the summation for k-dimensional group of positive integers (α_1, α_2,…, α_k)which satisfies this formula. In this note, our main purpose is to discuss computing problem of summation on equation (1).
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For any complex s, let ζ(s) denote Riemann Zeta function. We have ζ(s)=sum from n=1 to ∞ (1/n~s) when Re(s)>1. Now we define A(n,k,l)=sum from α_1+α_2+……+α_k=n to ((α_1α_2…α_k)~1ζ(2α_1)ζ(2α_2)…ζ(2α_k)), where n≥k is a positive integer, α_+α_2+…α_k=n denotes the summation for k-dimensional group of positive integers (α_1, α_2,…, α_k)which satisfies this formula. In this note, our main purpose is to discuss computing problem of summation on equation (1).
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